Question

Question 30 (4 points) Friedel-Crafts alkylation reactions cannot be performed on benzaldehyde because: The aldehyde group is strongly activating, and multiple substitutions occur on the aromatic ring. The aldehyde group gets alkylated instead of the aromatic ring. The aldehyde group attacks the Lewis acid to form an acid-base complex in which the aromatic ring is strongly deactivated The aldehyde group is moderately deactivating preventing any substitution on the aromatic ring. Question 31 (4 points) Which of the following produces no reaction? CHỊCONa + FCCOH CHệ Li*+ HC=CH- - O‘Na* + H20 CHỊCH Li* - HO

Answer 1

Solution:

Question 30:

Friedel-Craft reactions can not be performed on benzaldehyde because: The aldehyde group attacks the Lewis acid to form an acid base complex in which the aromatic ring is strongly deactivated.

0-AICI: + AlCl3 #

Question: 31

( C₆H₅-ONa + H₂O ) produces no reaction.

C₆H₅-ONa + H₂O will produce phenol and NaOH. And this NaOH (strong base) will remove an acidic proton from phenol and again it will produce C₆H₅-ONa (sodium phenoxide) + H₂O. So, overall there is no reaction.

ON CH ONa وه . . . . + H2O + NaOH + H2O

Therefore, it produces no reaction.

Question: 32

Option A is the correct product.

Step 1: Nitration, step 2: Chlorination, step 3: Reduction of nitro group, step 4: Diazotization, step 5: Sandmeyer reaction (CuCN / KCN) to give final product.

NO2 NO2 HNO3 H2SO4 Cl2 FeCl3 CI Fe HC1 CN N2C1 NH2 KCN CuCN/KCN CI NaNO2 HCl CI CI