Alcohols

The reaction between a monohydric alcohol (R–OH) and methylmagnesium iodide (an organometallic compound) can be represented as follows:

This reaction liberates methane (CH₄) gas as 3.1 mL. We need to find the molecular weight of the unknown alcohol.

To find the amount (in moles) of CH₄ gas evolved, use the Ideal Gas Law at standard temperature and pressure (STP):

1 mole of gas at STP occupies 22.4 L or 22400 mL.

Moles of CH₄ =

Moles of CH₄ =

The moles of R–OH are equivalent to the moles of CH₄ since 1 mole of R–OH generates 1 mole of CH₄ gas.

Given mass of R–OH = 4.5 mg = 0.0045 g

Molecular weight of R–OH =

Molecular weight =

Rounding to the nearest integer, the molecular weight of the unknown alcohol is 32 g/mol.

This calculated value of 32 g/mol does not fall within the 33-33 range, indicating potential non-STP conditions or additional factors not explicitly stated in the problem may affect this result. Verification of experimental conditions is advised.

The correct answer is 8

Note that phenyl ring has sp² carbons as well.

This is an example of a diazo coupling reaction, where a diazonium salt reacts with 1-naphthylamine to form an azo dye. The product is red in colour due to the extensive conjugation in the azo compound.