Hydrocarbons JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Hydrocarbons, free to read — no sign-in needed. The full chapter has 25 questions; sign in to attempt the remaining 22 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 1)
   Treatment of buta-1,3-diyne with $\text{NaNH}_2$ ($2$ equivalents), followed by reaction with excess of $trans$-$\text{CH}_3\text{-CH=CH-CH}_2\text{-Br}$ gives $\textbf{X}$ as the major product. The maximum number of carbon atoms that are collinear (in a straight line) in $\textbf{X}$ is ____.
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   Answer: 6

   Buta-1,3-diyne ($HC \equiv C - C \equiv CH$) reacts with $2$ equivalents of $\text{NaNH}_2$ to undergo double deprotonation, forming the acetylide dianion: $HC \equiv C - C \equiv CH + 2\text{NaNH}_2 \rightarrow \ ^\ominus C \equiv C - C \equiv C ^\ominus + 2\text{NH}_3$ This dianion acts as a strong nucleophile and undergoes an $S_N2$ reaction with the primary allylic halide, $trans$-$\text{CH}_3\text{-CH=CH-CH}_2\text{-Br}$ (excess ensures both sides react): $^\ominus C \equiv C - C \equiv C ^\ominus + 2 \text{CH}_3\text{-CH=CH-CH}_2\text{-Br} \rightarrow \text{CH}_3\text{-CH=CH-CH}_2 - C \equiv C - C \equiv C - \text{CH}_2\text{-CH=CH-CH}_3$ In the major product $\textbf{X}$, the central diyne unit consists of four $sp$ hybridized carbon atoms. The geometry around $sp$ hybridized carbons is linear (bond angle $180^\circ$). The two $sp^3$ hybridized $-\text{CH}_2-$ carbons attached directly to the $sp$ carbons also lie on this same linear axis. Thus, the sequence of collinear carbon atoms is: $-\text{CH}_2 - C \equiv C - C \equiv C - \text{CH}_2-$ The carbon atoms beyond the $-\text{CH}_2-$ groups are attached at a tetrahedral angle ($\sim 109.5^\circ$) and are not collinear with the central axis. Therefore, the maximum number of collinear carbon atoms in $\textbf{X}$ is $6$. Answer: $6$
2. Q2JEE Advanced Adv 2012 (Paper 1)
   In allene $\left(\mathrm{C}_{3} \mathrm{H}_{4}\right)$, the type(s) of hybridisation of the carbon atoms is (are):
   1. A.$s p$ and $s p^{3}$
   2. B.$s p$ and $s p^{2}$
   3. C.only $s p^{3}$
   4. D.$s p^{2}$ and $s p^{3}$
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   Answer: (B)

   Allene $\left(\mathbf{C}_{3} \mathbf{H}_{4}\right)$ is $\stackrel{s p^{2}}{\mathrm{H}_{2}} \mathrm{C}=\stackrel{s p}{\mathrm{C}}=\stackrel{s p^{2}}{\mathrm{CH}_{2}}$
3. Q3JEE Advanced Adv 2010 (Paper 1)
   The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are
   1. A.$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$$
   2. B.$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$$
   3. C.$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $$\mathrm{CH}_3 \mathrm{C} \equiv \equiv \mathrm{CH}$$
   4. D.$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$$a)
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   Answer: (D)

   $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2$ $-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ There are two (marked) possibility of halides Thus, in the given context $$\begin{array}{r} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}+\mathrm{NaNH}_2 \longrightarrow \\ \mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \overline{\mathrm{CNa}^{+}} \end{array}$$ $$\begin{aligned} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \overline{\mathrm{CNa}} \\ \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \\ \text { 3-octyne } \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \end{aligned}$$ Hydrocarbon Conceptual, reaction understanding II