What are X and Y respectively in the following set of reactions? (R--Br + \(CH_{3}CH_{2}O^-\) \(\rightarrow\) X; R--Br + \((CH_{3})_{3}CO^-\) \(\rightarrow\) Y)
Show Hint
Bulky bases usually favor elimination (\(E2\)), while small nucleophiles favor substitution (\(S_N2\)).
Step 1: Spot the deciding factor. A small strong nucleophile pushes a substitution reaction, while a big bulky base pushes an elimination reaction. The size of the attacking species matters most here. Step 2: Look at ethoxide ion. $CH_3CH_2O^-$ is a strong nucleophile and is not very bulky. Step 3: Predict product X. Because it is small, it attacks the carbon directly in an SN2 way and forms an ether (the substitution product). Step 4: Look at tert-butoxide ion. $(CH_3)_3CO^-$ is very bulky. It cannot reach the carbon easily, so it grabs a proton instead and acts as a base. Step 5: Predict product Y. Being bulky it favours E2 elimination and forms an alkene. Step 6: Match the option. The choice showing X as the ether and Y as the alkene is option 2. \[ \boxed{\text{Option 2}} \]
Was this answer helpful?
0
Top Questions on Nucleophilic and electrophilic substitution reactions (both aromatic and aliphatic)
