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Choose the correct option · Q1

Q.Which of the following represents the increasing order of boiling points of (1),

(2) and (3)?
(1) CH3−CH2−CH2−CH2−OHCH_3-CH_2-CH_2-CH_2-OH
(2) (CH3)2CH−O−CH3(CH_3)_2CH-O-CH_3
(3) (CH3)3COH(CH_3)_3COH
A.
(1) <
(2) <
(3)
B.
(2) <
(1) <
(3)
C.
(3) <
(2) <
(1)
D.
(2) <
(3) < (1)
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✓ Free question

Step 1. Identify each compound. (1) CH3-CH2-CH2-CH2-OH is n-butanol, a straight-chain primary alcohol. (2) (CH3)2CH-O-CH3 is isopropyl methyl ether, an ether of comparable molecular mass to (1) and (3). (3) (CH3)3C-OH is tert-butyl alcohol, a highly branched tertiary alcohol -- in fact a constitutional isomer of n-butanol, since both are C4H10O.

Step 2. Apply section 11.4.3/11.5.2's boiling-point rules. Alcohols hydrogen-bond strongly to each other (section 11.4.3), giving them much higher boiling points than an ether of similar molecular mass, which cannot hydrogen-bond to itself (section 11.5.2) -- so the ether (2) boils lowest of the three. Among the two alcohols, increased branching weakens intermolecular van der Waals contact and lowers boiling point (Problem 11.3): the more compact, more spherical tert-butyl alcohol (3) has LESS surface contact with neighbouring molecules than the elongated straight chain of n-butanol (1), so (3) boils below (1), even though both are alcohols of identical molecular formula.

Step 3. Combine and check against real data. n-Butanol boils at 118 degC, tert-butyl alcohol at 83 degC (Table 11.5), and a small ether of comparable mass (e.g. methyl isopropyl ether) boils well below both, around 30 degC. So the correct increasing order is (2) < (3) < (1) -- ether lowest, then the more branched (lower-boiling) alcohol, then the straight-chain (higher-boiling) alcohol.

✓Final answer

D. (2) < (3) < (1)

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