Straight-Line Carbon Lab

QMission brief — the question, verbatim

JEE (Advanced) 2026, Paper 1, Chemistry, Section 3 — the answer is a numerical value: +4 if only the correct value is entered, 0 in all other cases.

Q.12 Treatment of buta-1,3-diyne with NaNH₂ (2 equivalents), followed by reaction with excess of trans-CH₃-CH=CH-CH₂-Br gives X as the major product. The maximum number of carbon atoms that are collinear (in a straight line) in X is ____.

sp CARBONS

COLLINEAR C

The mystery. A rod-like diyne with an acidic hydrogen at each end, a strong base and an allylic bromide. Which bonds break, which form — and in the finished molecule, how far does the straight part reach? The lab watches both ends react, then spins X about its own axis: the carbons that never move are the ones on the line.

01Interactive experiment

One flask, watched at two levels at once — the flask on the left, the atoms on the right, in step. Press START EXPERIMENT: charge the flask → deprotonate each end → two SN2 attacks → hybridisation → the laser test → audit → answer.

Ready

01 Charge02 −H⁺03 −H⁺ again 04 SN205 SN2 again: X06 sp / sp² / sp³ 07 Laser test08 Audit09 Final answer

Set your values or use the question defaults.

Answer——

step 1 of 7

Two acidic ends, two equivalents of base, an excess of bromide. The question's reagents are loaded — press START EXPERIMENT.

✓ QUESTION VALUES LOADED

Customize experiment — optional

Instrument panel

step

—

temperature

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progress

0 %

NaNH₂ left

—

C⁻ ends

—

NaBr formed

—

arrows now

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sp · sp² · sp³

—

in one line

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Pink curved arrows move one electron pair each; red dashes are bonds breaking, green are bonds forming. The shape is built from the ideal angles sp 180°, sp² 120°, sp³ 109.5°.

Nothing needs to be chosen. Press START EXPERIMENT.

Bond angle at each carbon of the chain

from the hybridisation: sp 180°, sp² 120°, sp³ 109.5°

Distance of each carbon from the straight line

measured on the model — the carbons at 0 Å are the collinear ones

Result audit — X, the count and the traps

①X: both ends alkylated by SN2 (C₁₂H₁₄).

not yet examined

UNEVALUATED

②The sp carbons and their two neighbours: 6 in a line.

not yet examined

UNEVALUATED

③Trap: only the 4 sp carbons are collinear.

not yet examined

UNEVALUATED

④Trap: the line runs on through the CH₂ to 8.

not yet examined

UNEVALUATED

🧪Base Master

🏹Arrow Pusher

☂️Walden Flipper

📐Angle Reader

🪤Trap Spotter

🏆Chemistry Master

Reasoning log — input → action → change → observation → conclusion

Nothing investigated yet.

Evidence table (0)

023-D line explorer

Build a molecule from the same kind of reaction and spin it about its longest straight line. Atoms on the line never move; everything else swings round it. Change the alkyne, the amount of base or the bromide and see how the line grows or shrinks.

Alkyne

NaNH₂

Bromide

Spin about the line: 34°

03Detailed solution

The complete worked solution — every structure, step and calculation.

Understand the concept

Terminal alkynes are acids

An sp C–H (50 % s-character) has pKa ≈ 25; NH₃ has pKa ≈ 38, so NH₂⁻ removes it completely.

RC≡C–H + NH₂⁻ → RC≡C⁻ + NH₃

Acetylides are nucleophiles

The carbanion attacks a primary alkyl halide by SN2 — backside attack, one step, inversion at the CH₂.

RC≡C⁻ + R'CH₂Br → RC≡C–CH₂R' + Br⁻

sp carbons are linear

An sp carbon holds its two σ-bonded neighbours at 180°. A chain of sp carbons is straight — and so are the two atoms at its ends.

X–C≡C–C≡C–Y : all 6 on one axis

The line stops at sp³

The CH₂ is sp³ (109.5°): its next bond leaves the axis. The alkene beyond is sp² (120°).

answer = 4 sp C + 2 CH₂ = 6

Step 1 — Double deprotonation (2 equivalents of NaNH₂)

Each NH₂⁻ lone pair (gold dots) takes a terminal proton; the C–H bond pair stays on carbon. Both ends react because 2 equivalents are used.

The acetylide dianion ⁻C≡C–C≡C⁻ (2 Na⁺ counter-ions) and two molecules of NH₃ (pKa 38 ≫ 25, so the equilibrium lies fully to the right).

On the bench: the grey NaNH₂ was used up as the diyne was bubbled in; the proton went to NH₃ (the solvent).
Step 2 — SN2 alkylation at both ends (excess crotyl bromide)

SN2: the acetylide carbon attacks the CH₂ from the side opposite to Br (backside attack); the C–Br bond pair leaves with Br⁻. Transition state [C···CH₂···Br]⁻, trigonal-bipyramidal carbon; the CH₂ group inverts like an umbrella.

The same happens at the other end. Crotyl bromide is primary and allylic — ideal for SN2; with excess bromide both acetylide ends are alkylated. The C=C keeps its trans geometry because no bond to it is touched.

On the bench: bromide dripped in and white NaBr settled as Br⁻ met Na⁺.
Step 3 — The product X

X = (2E,10E)-dodeca-2,10-diene-5,7-diyne, CH₃CH=CHCH₂–C≡C–C≡C–CH₂CH=CHCH₃ (C₁₂H₁₄). Highlighted: the six carbons on one straight line.

⁻C≡C–C≡C⁻ + 2 CH₃CH=CHCH₂Br → CH₃CH=CHCH₂–C≡C–C≡C–CH₂CH=CHCH₃ + 2 Br⁻

Step 4 — Hybridisation of every chain carbon

Carbon (from one end)	Group	Hybridisation	Angle at this carbon	On the line?
C1	CH₃	sp³	—	no
C2, C3	CH=CH	sp²	120°	no
C4	CH₂	sp³	109.5°	yes — bonded to an sp carbon along its axis
C5–C8	–C≡C–C≡C–	sp	180°	yes
C9	CH₂	sp³	109.5°	yes
C10, C11	CH=CH	sp²	120°	no — the chain turns by 70.5° at C9
C12	CH₃	sp³	—	no

Step 5 — Count

C4 – C5 ≡ C6 – C7 ≡ C8 – C9 : angles at C5, C6, C7, C8 = 180° ⇒ all six on one axis
at C4 and C9 (sp³) the next bond turns off the axis by 180° − 109.5° = 70.5°
maximum collinear carbons = 4 (sp) + 2 (the CH₂ on each end) = —

Traps: counting only the sp carbons gives 4; letting the line run on past the CH₂ gives 8 or more; the trans C=C is not linear (120°).

Independent verification

RDKit templates. Deprotonation and SN2 templates applied to buta-1,3-diyne and trans-crotyl bromide give X = C₁₂H₁₄ with the E,E geometry kept.
3-D geometry. X embedded and optimised with MMFF: the largest set of carbons within 0.1 Å of one straight line is 6.
Controls. Hexatriyne gives 8, ethyne 4, one equivalent of base 5, allyl or propargyl bromide 6 — the same rule every time.
Solution websites. MathonGo's published solution (their Q44) gives 6.
This page. The arrows are pushed on the molecule graph; hybridisation is read from bond orders; the count is made twice (by rule and on the geometric model) at run time.
Answer key. Checked afterwards against the official final key printed with the paper: Q12 → 6.

04How to use this simulation

For a student opening it for the first time. The whole investigation runs by itself after one press.

STEP 1

Read the mission brief

A diyne, two equivalents of NaNH₂ and excess crotyl bromide give X. How many carbons of X are in one straight line?

STEP 2

Check the values

The question's reagents are loaded. You do not need to change anything. The answer is shown above the experiment.

STEP 3

Optional: customize

Try a triyne or ethyne, one equivalent of base, or another bromide. The lab says honestly when you have left the question. ↺ RESTORE QUESTION CONDITIONS puts everything back.

STEP 4

Start

Press ▶ START EXPERIMENT under the experiment screen. Nothing else to click (about a minute; SLOW MOTION or 2× changes the pace).

STEP 5

Watch both levels

Left: the ammonia flask — grey NaNH₂ used up, bromide dripping, NaBr settling. Right, at the same moment: curved arrows, bonds breaking and forming, the carbanions and the SN2 umbrella flip.

STEP 6

The laser test

X relaxes into its real angles and spins about its straight axis; the carbons that stay still are counted. Badges unlock on the way.

STEP 7

Explore

In the 3-D line explorer build other products and spin them. Then read the Detailed solution.

STEP 8

Experiment again

Press ▶ REPLAY EXPERIMENT, or change a reagent and press ⚙ RUN WITH CUSTOM VALUES. CLASSROOM MODE enlarges everything for a projector.

Classification

Exam	JEE Advanced 2026, Paper 1, Chemistry, Section 3, Q.12
Chapter	Hydrocarbons
Topic	Alkynes - acidity of terminal alkynes, acetylide SN2 alkylation and linear sp geometry
Difficulty	Easy-Moderate
Answer	— run the experiment

Straight-Line Carbon Lab — how many carbons of X lie on one line?

QMission brief — the question, verbatim

01Interactive experiment