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# Organic Chemistry Question ? Watch

1. A branched Alkene A was reduced with Hydrogen to form a compound B. Compound B had a composition by mass of C:82.8%, H:17.2% (Mr=58).

Compound A was reacted with steam with an acid catalyst to produce a mixture of two structural isomers C and D .

a)Calculate the empirical and molecular formula of a compound B?

I'd like some advice as to where to start trying to work out how to tackle this question.

any help is greatly appreciated .

kind regards to anyone who is helping me
2. The reaction of an alkene and hydrogen is an electrophilic addition which forms an alkane. As we know this alkane has mass 58, we can deduce that it is an isomer of butane, which can be reinforced by the percentage compositions of carbon and hydrogen. Therefore, compound A is an isomer of Butene
3. To add to my previous answer, now that we know alkene A is an isomer of Butene, we can deduce compounds C and D, as the hydration of an alkene will form 2 alcohols; a major and minor isomer of Butanol
4. (Original post by phspa)
A branched Alkene A was reduced with Hydrogen to form a compound B. Compound B had a composition by mass of C:82.8%, H:17.2% (Mr=58).

Compound A was reacted with steam with an acid catalyst to produce a mixture of two structural isomers C and D .

a)Calculate the empirical and molecular formula of a compound B?

I'd like some advice as to where to start trying to work out how to tackle this question.

any help is greatly appreciated .

kind regards to anyone who is helping me
An alkene reduced with hydrogen forms an alkane so that is your compound B

To calculate empirical formula you divide the % composition by the elements atomic mass (for example 12 for carbon) which should give you two answers. Further divide these figures by the smallest one which gives you the empirical formula. For the molecular formula you already know its an alkane which has the general formula CnH2n+2.

I do believe an alkene reacting with steam and acid gives a primary alcohol but I could be wrong
5. Furthermore, as Butene only has one branched isomer: 2-Methyl Propene, we can assume that this is compound A, so compounds C and D must be 2-Methyl propan-1-ol and 2-methyl propan-2-ol, as these are the major and minor products of the hydration of 2-Methyl Propene
Sorry for the various posts, but I couldn't remember the question whilst I wrote the whole thing

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