Question

Which of the following four compounds (I to IV) are correctly arranged in decreasing order of reactivity towards \(S_{N}2\) reaction?
I. 1-Bromobutane
II. 1-Bromo-2-methylbutane
III. 1-Bromo-2,2-dimethylpropane
IV. 1-Bromo-3-methylbutane

• A. \(I > IV > III > II\)
• B. \(I > II > III > IV\)
• C. \(I > III > IV > II\)
• D. \(IV > III > II > I\)

Hint:

\(S_N2\) reactions are extremely sensitive to steric hindrance. Branching near the reactive carbon slows the reaction drastically.

Answer 1

The Correct Option is A

Step 1: Recall how SN2 works.
In an SN2 reaction the nucleophile attacks the carbon from the back. Anything that crowds that carbon slows the reaction. So less crowding means faster reaction.
Step 2: Look at compound I.
1-Bromobutane is a straight chain with no branch near the reacting carbon, so it is the least crowded and reacts fastest.
Step 3: Look at compound IV.
1-Bromo-3-methylbutane has its branch far from the reacting carbon, so it is only slightly slowed.
Step 4: Look at compound III.
1-Bromo-2,2-dimethylpropane (neopentyl type) has heavy crowding fairly close, so it reacts slower than IV.
Step 5: Look at compound II.
1-Bromo-2-methylbutane has a branch right next to the reacting carbon, so it is the most blocked and slowest.
Step 6: Order them.
\[ \boxed{I>IV>III>II} \]