Question 3
NECO
•
2019
•
CHEMISTRY
•
objective
Two primary alcohols with molecular formula \(\mathrm{C_4H_{10}O}\) have boiling points of 108°C and 117°C respectively. Identify the structural formulas of these alcohols and explain the difference in their boiling points.
Answer Options
A.
1‑butanol: CH3CH2CH2CH2OH (b.p. 117°C); 2‑methyl‑2‑propanol: (CH3)3COH (b.p. 108°C). The branched isomer boils lower due to reduced surface area.
Correct Answer
B.
1‑butanol: CH3CH2CH2CH2OH (b.p. 117°C); 2‑methyl‑1‑propanol: (CH3)2CHCH2OH (b.p. 108°C). The branched isomer boils lower due to reduced surface area and weaker intermolecular forces.
C.
1‑butanol: CH3CH2CH2CH2OH (b.p. 117°C); 2‑methyl‑1‑propanol: (CH3)2CHCH2OH (b.p. 108°C). The difference arises from the presence of a carbonyl group in the branched isomer.
D.
1‑butanol: CH3CH2CH2CH2OH (b.p. 108°C); 2‑methyl‑1‑propanol: (CH3)2CHCH2OH (b.p. 117°C). The linear isomer boils lower because it has a smaller molecular weight.
Correct Answer
B
Correct Option:
1‑butanol: CH3CH2CH2CH2OH (b.p. 117°C); 2‑methyl‑1‑propanol: (CH3)2CHCH2OH (b.p. 108°C). The branched isomer boils lower due to reduced surface area and weaker intermolecular forces.
About This Question
This is
Neco
2019
Chemistry
Question 3.
It is one of the
objective
questions from the
2019
Neco
Chemistry
examination.
Difficulty level:
Medium
.