An iodoalkane A, with a molecular mass of 198, contains 30.3% carbon and 5.6% hydrogen by mass. It reacts with aqueous sodium hydroxide solution to give an alcohol which can be oxidised and with ethanolic potassium hydroxide solution to give two alkenes, neither of which show geometrical isomerism. Identify A and show the elimination reaction and state the products.

I got 2-iodo-3-methylbutane for the first one but the second one I am a bit confused as to what the product looks like and where the hydroxide reacts with the hydrogen

jhk

To identify the iodoalkane A, we can start by calculating the number of carbon and hydrogen atoms in A based on the given percentages.

Let's assume we have 100 grams of A:
- The mass of carbon in A is 30.3 grams (30.3% of 100 grams).
- The mass of hydrogen in A is 5.6 grams (5.6% of 100 grams).

Next, we need to convert the mass of carbon and hydrogen to moles using their respective atomic masses.
- The atomic mass of carbon (C) is 12.01 g/mol, so the number of moles of carbon is 30.3 grams / 12.01 g/mol.
- The atomic mass of hydrogen (H) is 1.01 g/mol, so the number of moles of hydrogen is 5.6 grams / 1.01 g/mol.

From the balanced chemical equation of the reaction that forms the alcohol, we know that an iodoalkane reacts with sodium hydroxide (NaOH) to produce an alcohol and sodium iodide (NaI). The general reaction can be represented as:
RI + NaOH → ROH + NaI

Since A is an iodoalkane, it contains an iodine atom (I). The molecular mass of A is given as 198 g/mol, so the remaining mass is due to the iodine atom.
- The molecular mass of iodine (I) is 126.9 g/mol, so the mass of iodine in A is 198 grams - (mass of carbon + mass of hydrogen).

By calculating the number of moles of C, H, and I, we can determine their respective molecular formulas and structures.

Now let's move on to the reaction with ethanolic potassium hydroxide (KOH) to form two alkenes. This is an elimination reaction, specifically an E2 reaction. In an E2 reaction, the hydroxide (OH) attacks the hydrogen on the beta-carbon (the carbon attached to the iodine atom). The iodine and hydrogen atoms are eliminated as a leaving group, forming a double bond between the alpha- and beta- carbons.

The reaction can be represented as:
2-iodo-3-methylbutane (A) + KOH → Alkene 1 + Alkene 2 + KI + H2O

The specific structures of the resulting alkenes depend on the position of the beta-carbon (the carbon directly attached to the iodine atom) in iodoalkane A. It would be helpful to know the structure of A to determine the exact structures of the alkenes formed.

Do you have the structure or the name of iodoalkane A?