Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Alcohols, Phenols and Ethers

Multiple Choice Questions

91.

( +2) 2-methylbutan-1-ol (-) 2-methylbutan -1- ol have different values for which property?

Boiling point
Relative density
Refractive index
Specific rotation

Specific rotation

( +) 2-methyl butan-1-ol and (-) 2-methyl butan-1-ol are enantiomer. Enantiomers are non-superimposable mirror images of each other hence, are optically active.

The optical activity of a compound can be confirmed by the value of specific rotation. It is defined as the rotation produced by solution of 10 cm length and unit concentration (1 g/ mL) for the given λ of light at the given temperature.

92.

375 mg of an alcohol reacts with required amount of methyl magnesium bromide and releases 140 mL of methane gas at STP. The alcohol is

ethanol
n-butanol
methanol
n-propanol

n-propanol

The required reaction is

$\underset{375 mg}{ROH} + {CH}_{3} MgBr \to \underset{140 mL at STP}{{CH}_{4}} + Mg (OR) (Br)$

Moles of CH₄(n) at STP = $\frac{140}{22400}$ = 0.00625 moles

Here, -CH₃ of CH₃MgBr combines with H-atom of R-OH to give CH₄.

$∴$ n(CH₄) = n(R-OH)

Also, n = C_nH_2n+1.OH

Thus, 12n + 2n + 1 + 17 = 60 ( $∵$ OH = 16 + 1 = 17)

$∴$ 14n = 60 - 18 = 42

n = 3

Hence, alcohol is n-propanol (C₃H₇OH).

93.

Phenol is a highly corrosive substance, but it 0.2 percent solution is used as

Antibiotic
Antiseptic
Disinfectant
Antihistamine

Antiseptic

0.2% phenol solution is used as an antiseptic solution.

94.

The compound that would produce a nauseating smell/ odour with a hot mixture of chloroform and ethanolic potassium hydroxide is

PhCONH₂
PhNHCH₃
PhNH₂
PhOH

PhNH₂

When primary amine reacts with chloroform and ethanolic potassium hydroxide, they produce isocyamides as main product. The reaction is also known as carbylamine reaction.

Ph-NH₂ $\underset{(Nauseating smell)}{\to_{KOH (ethanolic)}^{{CHCl}_{3}}}$ Ph.N → C

95.

Reduction of the lactol S OH with sodium borohydride gives

The reaction occurs as follows:

96.

The main product (P) is

C₆H₅OC₂H₅
C₂H₅OC₂H₅
C₆H₅OC₆H₅
C₆H₅.I

C₂H₅OC₂H₅

Since, C₆H₅O^- is a weaker nucelophile than C₂H₅O^-, therefore, the stronger nucleophile, i.e. C₂H₅O^- will attack on C₂H₅I to give diethyl ether.

(i) C₆H₅OH + C₂H₅O^- → C₆H₅O^- + C₂H₅OH

(ii) C₂H₅O^- + CH₃.CH₂.I → CH₃CH₂-O-CH₂CH₃

97.

Name of the following reaction is

Reimer-Tiemann
Kolbe-Schmitt
Cannizzaro
Gattermann

Kolbe-Schmitt

The given reaction is known as Kolbe-Schmidt reaction. It is used to get salicylic acid.

98.

The order of reactivity of the following alcohols is

1 > 2 > 3 > 4
1 > 3 > 2 > 4
4 > 3 > 2 > 1
4 > 3 > 1 > 2

4 > 3 > 2 > 1

The order of reactivity depends on the stability of the carbo cations of the respective species. The carbocations formed are

F.CH₂. $\overset{+}{C}$ H.CH₃, FCH₂.CH₂ $\overset{+}{C}$ H.CH₃

CH₃. $\overset{+}{C}$ H.CH₃and Ph. $\overset{+}{C}$ H₂

The stability order of carb-cation is

Ph $\overset{+}{C}$ H₂ > CH₃. $\overset{+}{C}$ H.CH₃ > F.CH₂.CH₂. $\overset{+}{C}$ H.CH₃ > F.CH₂.CH₂. $\overset{+}{C}$ H.CH₃ > FCH₂ $\overset{+}{C}$ H.CH₃

Hence, order of reactivity follows the order (4) > (3) > (2) > (1)

99.

Which of the following compound (s) will give butanone on oxidation with alkaline KMnO₄ solution?

Butan-1-ol
Butan-2-ol
Both (a) and (b)
None of the above

Butan-2-ol

Alkaline KMnO₄ is a strong oxidising agent. It oxidises primary alcohols to aldehyde and secondary alochols to ketones.

(i) $\underset{(1 °)}{{CH}_{3} {CH}_{2} {CH}_{2} {CH}_{2}} - OH \to_{{KMnO}_{4} (alkl .)}^{[O]} \underset{Butanol}{{CH}_{3} {CH}_{2} {CH}_{2} {CH}_{3}}$

(ii) $\underset{(2 °)}{{CH}_{3} {CH}_{2}} \underset{OH}{\underset{|}{CH}} {CH}_{3} \to_{{KMnO}_{4} (alk .)}^{[O]} \underset{Butanone}{{CH}_{3} {CH}_{2} \underset{O}{\underset{| |}{C}} {CH}_{3}}$

100.

What are the organic products formed in the following reaction?

$C_{6} H_{5} - COO - {CH}_{3} \to_{(ii) H_{2} O}^{(i) {LiAlH}_{4}}$

$C_{6} H_{5} - COOH and {CH}_{4}$
$C_{6} H_{5} - {CH}_{2} - OH and {CH}_{4}$
$C_{6} H_{5} - {CH}_{3} and {CH}_{3} - OH$
$C_{6} H_{5} - {CH}_{2} - OH and {CH}_{3} - OH$

$C_{6} H_{5} - {CH}_{2} - OH and {CH}_{3} - OH$