Nitro group is an electron-withdrawing group and will increase the strengths of acids.

All India 2015)

2 Answers. Question 98. (i) There are two – NH2 groups in semicarbazide. (b) (i) Molecular formula of compound : C9H10O (i) Acetylene to Acetic acid (ii) Toluene to m-nitrobenzoic acid Answer: Question 33. Benzaldehyde reduces Tollen’s reagent to give a red-brown precipitate of Cu20, but aceptophenone being a ketone does not. Write structural formulae and names of the four possible aldol condensation products from propanal and butanal. (a) Account for the following : These are known as methyl ketones. (All India 2012) (a) Give chemical tests to distinguish between. (ii) Ph — CH = CH — CHO : (a) State and illustrate the following : reagent and undergoes Cannizzaro’s reaction. Delhi 2017)

Formaldehyde does not take part in Aldol condensation. Aldol condensation: (ii), (v), (vi), (vii). Write the structures of A, B, C, D and E in the following reactions : (Delhi 2016)

(Comptt.

(b) (i) Ethanal and Propanal : Ethanal and propanal can be distinguished by iodoform test.

Answer:

of iodoform but benzaldehyde does not. Expert Answer 100% (1 rating) Previous question Next question Get more help from Chegg. In each case, indicate which aldehyde served as nucleophile and which as electrophile.

By Iodoform test : Acetophenone being a methyl ketone on treatment with I2/NaOH (NaOI) undergoes iodoform test to give yellow ppt.

Answer:

Answer: (a) Write the products formed when CH3CHO reacts with the following reagents :

Question 106.

(v) On oxidation it gives a mixture of ethanoic acid and propanoic acid Answer: RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, Important Questions for Class 12 Chemistry, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. Answer:

(i) Propanone to Propan-2-ol (ii) Ethanal to 2-hydroxy propanoic acid

(a) (i) Propanone gives iodoform test while propanol does not give this test.

On vigorous oxidation, it gives ethanoic acid and propanoic acid. Do the following conversions in not more than two steps:

Question 13. (Comptt.

Iodoform test: Warm each organic compound with I2 and NaOH solution. Question 57.

(v) It does not decolorise Br2 water or Baeyer’s reagent so the unsaturation of ‘A’ is due to benzene ring.

Benzaldehyde does not react. (ii) Cannizzaro reaction.

(a) Write the IUPAC names of the following compounds : (iii) Benzaldehyde and Acetophenone (Comptt.

Its functional isomer A will be propanal. (ii) Hell-Volhard-Zelinsky reaction : Carboxylic acid reacts with chlorine or bromine in presence of small quantities of red phosphorous to give exclusively a-chloro or a-bromo acids. Ethanal is the only aldehyde to give the triiodomethane (iodoform) reaction. (i) Propanol and propanone (ii) Benzaldehyde and acetophenone (ii) Carboxylic acid is a stronger acid than phenol.

Write the IUPAC name of the following: (Comptt.

All India 2013) Click the test you would like to run on benzaldehyde. aqueous alkali giving product known as Aldol. (b) Complete each synthesis by giving missing reagents or products in the following : (All India 2011)

∴ A = Formaldehyde, B = Methyl alcohol, C = Sodium formate, D = Formic acid. (b) Give chemical tests to distinguish between the following pairs of compounds : (i) Benzoic acid and Phenol (ii) Propanal and Propanone.

(i) In the resonating structure of phenol and carboxylic acid, the negative charge on the carboxylate ion is delocalised over two oxygen atoms while they are localized on oxygen atom. Write the structure of 3-oxopentanal. (b) Distinguish between: 2,4-DNP Test for Aldehydes and Ketones.

On vigorous oxidation it gives ethanoic and propanoic acids. Question 83. (a) (i) Acetylation: The addition of acyl (-COCH3) group on the ortho and para positions of acyl halide in the presence of a Lewis acid i.e.

Benzophenone does not respond to this test. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test.

The products are: Cannizzaro reaction: (i), (iii), (ix). Warm each compound with iodine and sodium hydroxide solution in water. One of its functional isomer i.e., B shows iodoform test which can be only shown by compounds having methyl ketone so the compound B will be Acetone or 2-propanone. Warm each compound with iodine and sodium hydroxide solution in water. All India 2014) Example : Write IUPAC name of the following : (Comptt. IUPAC name : 3-Hydroxybutanoic acid, Question 28.

Find out more about how we use your information in our Privacy Policy and Cookie Policy. Aldehydes and Ketones have lower boiling points than corresponding alcohols. (b) (i) Ethanal and Propanal : Ethanal and propanal can be distinguished by iodoform test.

(Comptt. endstream endobj startxref Question 17. After losing a proton, carboxylic acids gain a negative charge as shown: Now, any group that will help stabilise the negative charge will increase the stability of the carboxyl ion and as a result, will increase the strength of the acid.

(ii) Carboxylic acids do not give characteristic reactions of carbonyl group. (a) Write the products of the following reactions : (e) Propanal and Propanone: Propanal being an aldehyde gives positive test with Fehling solution in which a red brown ppt. (ii) Friedel-Crafts reaction: The introduction of alkyl or acetyl group in the presence of anhydrous almuninium chloride (AlCl3) as catalyst to ortho and para positions of an aromatic compound is called Friedel-Crafts reaction. Delhi 2014)

Answer:

Does benzaldehyde give a positive or negative result for an iodoform test.

Answer: On vigorous oxidation, it gives 1,2- benzenedicarboxylic acid.

7 years ago. Derive the possible structure of compound A.

(a) (i) Etard reaction.

Draw the structure of 2-methylbutanaI. Acetophenone, being a methyl ketone on treatment with I2/NaOH undergoes iodoform reaction to give a yellow ppt.

(b) An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. Delhi 2012)

Question 97.

(i) HCN (ii) H2N – OH (iii) CH3CHO in the presence of dilute NaOH

(a) (i)Phenol and Benzoic acid: On addition of NaHCO3 to both solutions carbon dioxide gas is evolved with benzoic acid while phenol does not form CO2 (b) An organic compound with molecular formula C9H10O forms 2, 4-DNP derivative and reduces 237 0 obj <> endobj Answer: (b) An organic compound with molecular formula C5H10O does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. (a) Illustrate the following name reactions giving a chemical equation in each case : Answer: Answer: Question 53.

Question 4. (i) Ethanal and Propanal (a) How will you carry out the following conversions? Give a chemical test to distinguish between Benzoic acid and Phenol.

(b) An organic compound A has the molecular formula C8H16O2. All India 2012)

(i) Etard reaction (ii) Stephen reduction Question 99. Answer: Question 21. Question 64. Benzaldehyde reduces Tollen’s reagent to give a red-brown precipitate of Cu20, but aceptophenone being a ketone does not. of cuprous oxide is obtained while propanone being a ketone does not respond to this test.

Answer: It can be observed that steric hindrance also increases in the same. (iii) Bromobenzene to benzoic acid (All India 2010) (Delhi 2014) (ii) Etard reaction. Answer:

(ii) Clemmensen reduction: The carbonyl group of aldehyde and ketones is reduced to CH, group on treatment with zinc amalgam and concentrated hydrochloric acid. (ii) The release of a proton from carboxylic acids is much easier than from phenols,

Butanal gives silver mirror whereas Butan-2-one does not. Draw the structure of 4-chloropentan-2-one.

is a polar group in which carbon acquires positive charge and O acquires negative charge due to more electronegativity of oxygen.

Reactions : Question 74.

(ii) Phenol and Benzoic acid: On addition of NaHCO3 to both solutions carbon dioxide gas is evolved with benzoic acid while phenol does not form CO2 All India 2012) (b) (i) Wolff-Kishner reduction reaction : The reduction of aldehydes and ketones to the corresponding hydrocarbons by heating them with hydrazine and KOH or potassium tert-butoxide in a high boiling solvent like ethylene glycol is called Wolff-Kishner reduction. Standards Cyclohexanone, Benzophenone, and Benzaldehyde. (iii) Phenol and Benzoic acid (Delhi 2012) Answer: (b) Which acid of each pair shown here would you expect to be stronger? Question 16.

(Comptt.

Ethanal is soluble in water due to H-bonding between the polar carbonyl group and water molecules. Answer: Question 2.

(i) Clemmensen reaction (ii) Cannizzaro’s reaction (a) (i) Cannizzaro’s reaction: Aldehydes, which do not have an oc-hydrogen atom undergo self oxidation and reduction on treatment with cone, alkali and produce alcohol and carboxylic acid salt. (a) Give chemical tests to distinguish between the following pairs of compounds : (i) Aldol condensation (ii) Cannizzaro reaction (iii) On adding NaHCO,, benzoic acid produces brisk effervescence of C02 gas whereas ethylbenzoate does not. Question 54. Arrange the following compound groups in the increasing order of their property indicated: Propanol, Propane, Propanal (boiling point) (Delhi 2017)

Delhi 2015) (ii) Cl—CH2—COOH, F—CH2—COOH, CH3—COOH (acidic character) (All India 2015) (i) Pentan-2-one and Pentan-3-one (ii) Ethanal and Propanal (All India 2013) (b) (i) Pollen’s reagent test. Question 62.

(i) Ethanol to 3-hydroxybutanal (ii) Benzaldehyde to Benzophenone

(b) An organic compound with molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollen’s

Question 22. Answer: Question 7. On addition of NaHCO3 to both solutions carbon dioxide gas is evolved with benzoic acid while phenol does not form CO2.

Arrange the following compounds in increasing order of their property as indicated : (Comptt. Delhi 2013) (i) Propanal and Propanone, (ii) Benzaldehyde and Acetophenone. (b) Write the structures of the main products of following reactions : (Delhi 2012) (ii) Benzaldehyde and acetophenone : But benzaldehyde does not respond to this test. Answer.

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