Chemistry MCQ thread...

[Zishi]

N09 paper 11 Q 28?

The answer should be C. According to the nucleophilic reaction mechanism, first of all a negatively charged thing(nucleophile) will attach the carbon atom of carbonyl group. Oxygen will take an electron from that carbon atom and will get a negative charge. Then the hydrogen from CH3CO2CH3 will be attached with that negatively charged oxygen. But be careful, you shouldn't change the hydrocarbon tails of the ester group.I mean that carbon of ester group should remain attached to CH2 and oxygen from ester group should remain attached to methyl group. This way only one option seems to be the answer. That is C.

 

[Anonymousx3]

What effect does KMnO4 have on alcohols and esters?
The answer's B, by the way, but I don't understand how. :\

 

[Anonymousx3]

Why is the answer C? D:

 

[aquariangurl]

Primary alcohols give aldehydes, secondary give ketones & tertiary do not change on oxidation with Kmno4
In this case, B is a tertiary alcohol
& it also releases H2 in a reaction with Na, The COH becomes CONa + H, Got it?

 

[Zishi]

What effect does KMnO4 have on alcohols and esters?
The answer's B, by the way, but I don't understand how. :\

Primary and secondary alcohols can be oxidised to their respective carbonyl compounds by oxidation with potassium permanganate or potassium dichromate. B is a tertiary alcohol which can't be oxidised. Alcohols also give hydrogen gas when reacted with sodium metal. Esters can't be oxidised and they also don't give +ve test with sodium metal. So the answer is B.

 

[Zishi]

Why is the answer C? D:

Look at the second molecular formula carefully. There's an increase in two oxygen atoms. It means that two carboxylic acid groups have been formed on oxidation. And they can only be formed by aldehydes(primary alcohols). And aldehyde functional groups can only be present at the end of the chains.

 

[Anonymousx3]

The answer's C, but shouldn't it be B?
There's already one ketone group and then you get another on oxidation, which is 2.
Or does the other C=C bond in the benzene ring have something to do with it?

 

[Zishi]

The answer's C, but shouldn't it be B?
There's already one ketone group and then you get another on oxidation, which is 2.
Or does the other C=C bond in the benzene ring have something to do with it?

In the very first double bond, CH2 will give you carbon dioxide and water. On double bond's other side there's a methyl group and a ring(consider it as R). As in this case double bonded carbon atom has two other carbon atoms(Or groups) attached to it, so it will be oxidised to a ketone. And for the double bond in the ring. One carboxylic acid will be formed, plus one ketone after the oxidation of that double bond. That leaves you with three ketonic groups only. So you'd need 3 2,4-DNPH molecules. Also, that IS NOT a benzene ring.

 

[Anonymousx3]

'kayy, last question! (for now xD).
I get 38% when I do this. My calculation is:

1 mol ethanol gives 1 mol ester
46g ethanol gives 88g ester
so 30g of ethanol should give (30/46) x 88 = 57.4g ester
However, 22g are actually formed so the %age yield = (22/57.4) x 100 = 38%

The actual answer is 50%.

 

[Zishi]

'kayy, last question! (for now xD).
I get 38% when I do this. My calculation is:

1 mol ethanol gives 1 mol ester
46g ethanol gives 88g ester
so 30g of ethanol should give (30/46) x 88 = 57.4g ester
However, 22g are actually formed so the %age yield = (22/57.4) x 100 = 38%

The actual answer is 50%.

No, you've to do this using moles. CIE has given the same masses because they wanted to make the question misleading. Make moles of both ethanol and ethanoic acid. Use the least number of moles to form moles of ester. Those moles of ester will be same as that least number. Convert those moles into grams of the ester. Then divide 22 by that mass. That should give you the correct answer. The reason behind it is that one of them is a limiting reagent.

 

[Anonymousx3]

Thanks Zishi, you really helped me out!
Good luck with any exams you have left!

 

[Zishi]

Thanks Zishi, you really helped me out!
Good luck with any exams you have left!

No problem. And thanks for the good luck - I really need it.

 

[hassam]

so ready for a tough one ZiShi....

 

[hassam]

An element,X, conducts electricity in the solid. It has a very high melting point. Its chloride does not dissolve in water. One of its oxides dissolves moderately in water to give a faintly acidic solution.X is likely to be
A aluminium
B graphite
C magnesium
D lead

 

[aquariangurl]

^A?

 

[MHHaider707]

Its B i-e Graphite!!!

 

[aquariangurl]

Can you tell me WHY?

 

[MHHaider707]

so ready for a tough one ZiShi....

C is the right one!!!

 

[Xthegreat]

all A, B, C and D have high melting point.

- graphite is an allotrope of carbon ( meaning the same element of different form )
- Aluminium chloride dissolves in water to give a very acidic solution / magnesium chloride dissolves partially water / chloride of lead and CCl4 do not dissolve in water
- aluminium oxide is insoluble in water / magnesium oxide and lead (II) oxide dissolve in water to give alkaline solution / CO2 dissolves partially in water to give an acidic solution.

ok?

 

[MHHaider707]

Can you tell me WHY?

See it says on dissolving in water oxide of X gives acidic solution so graphite is the answer!!!

Aluminium can't be the answer coz its oxide is amphoteric i-e it reacts with base like acid and with acid like base!!!


got it??