The gas evolved on heating $$CH_3MgBr$$ in methanol is:

First, we recall a very important general fact about Grignard reagents. A Grignard reagent has the general formula $$R{-}MgX$$, where $$R$$ is an alkyl group and $$X$$ is a halogen. Because the carbon-magnesium bond is highly polar, the carbon atom behaves as a strong carbanion. Therefore, a Grignard reagent is extremely reactive toward any source of acidic hydrogen (denoted by $$H^+$$), such as water, alcohols, phenols, carboxylic acids, and even terminal alkynes.

The standard reaction that expresses this reactivity is written first in its most general form:

$$R{-}MgX \;+\; H{-}Y \;\rightarrow\; R{-}H \;+\; MgX{-}Y$$

Here, $$H{-}Y$$ is any compound supplying a proton. The key observation is that the $$R$$ group from the Grignard reagent simply abstracts (removes) the proton. As a result, the hydrocarbon $$R{-}H$$ is liberated, often as a gas if it is small and volatile, while the magnesium-containing residue $$MgX{-}Y$$ remains in solution or as a solid.

Now we apply this general idea to the specific substances given in the problem. The Grignard reagent supplied is $$CH_3MgBr$$. The proton source present is methanol, which we write as $$CH_3OH$$. Placing these into the general reaction pattern, we write:

$$CH_3MgBr \;+\; CH_3OH \;\rightarrow\; CH_3H \;+\; MgBr(OCH_3)$$

At this stage we notice that the formula $$CH_3H$$ can be simplified. Joining the carbon atom with three hydrogens from the methyl group and the single hydrogen abstracted from methanol gives the simplest alkane, methane, whose correct molecular formula is $$CH_4$$. Therefore we rewrite the equation more neatly as:

$$CH_3MgBr \;+\; CH_3OH \;\rightarrow\; CH_4 \;+\; Mg(OCH_3)Br$$

Because methane is a very small, non-polar molecule, it is a gas under ordinary laboratory conditions. Consequently, when the mixture is warmed, methane escapes as a gas.

We can underline the reasoning in words: the methyl group $$CH_3{-}$$ from the Grignard reagent accepts a proton $$H^+$$ from methanol, directly forming methane $$CH_4$$. No other gaseous hydrocarbon (such as ethane or propane) can appear, because only one carbon atom is present in the fragment leaving as the hydrocarbon.

Therefore, the gas evolved is methane, which corresponds to Option A.

Hence, the correct answer is Option A.