Chemistry: Post your doubts here!

[mittul8yu98]

very very urgent help

 

[student3737]

Why doesn’t the second reaction occur ?

 

[blymphocytes]

the answer is so stupid

These structures look so ugly....and how are we even supposed to know thattt -_-

 

[blymphocytes]

Why doesn’t the second reaction occur ?

What are these stupid qns??

 

[blymphocytes]

Why doesn’t the second reaction occur ?

Ok i think it may be because the electrode potentials dont have a large difference as compared to the previous one.
I'm not sure though

 

[A*****]

For biii) why do we divide the moles we got in bii) with 3?

From 1 mole of acid we are getting 1 mole of SO2 and 1 mole of HCl. For 1 mol of SO2, 2 moles of NaOH and for 1 mole of HCl 1 mole of SO2 is required. Therefore, to neutralize the products of 1 mole of acid, 3 moles of NaOH are required. Now apply simple ratio to get the answer

 

[A*****]

very very urgent help

What is the ans?

 

[mittul8yu98]

What is the ans?

Dont know man just send ur answers

 

[MShaheerUddin]

Can anyone tell what are the question that examiner in p4 usually asks about chromatography except RF and circling.

 

[oalevel2020]

all the topics in the syllabus bt that are sum notes to be learnt easily in pdf file

Can you please send me the notes?
Thank you in advance.

 

[mittul8yu98]

Why doesn’t the second reaction occur ?

Because both specues are reduced

 

[ba-lock-ey]

These structures look so ugly....and how are we even supposed to know thattt -_-

A condensation reaction occurs, so naturally if we have the same number of carbons as before, it's like a self condensing reaction so I suppose that's how the ring came to be.

"Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine." To quote

As for the second part, LiAlH4 is a strong reducing agent as it isn't in dry ether so the c=o becomes ch2. FYI RCO2H and LiAlH4 gives an alcohol but Carbonyl groups give the alkane right away with this.

 

[blymphocytes]

A condensation reaction occurs, so naturally if we have the same number of carbons as before, it's like a self condensing reaction so I suppose that's how the ring came to be.

"Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine." To quote

As for the second part, LiAlH4 is a strong reducing agent as it isn't in dry ether so the c=o becomes ch2. FYI RCO2H and LiAlH4 gives an alcohol but Carbonyl groups give the alkane right away with this.

Thank you so muchhh

 

[studyingrobot457]

Good luck for tmrw guys

 

[mittul8yu98]

A condensation reaction occurs, so naturally if we have the same number of carbons as before, it's like a self condensing reaction so I suppose that's how the ring came to be.

"Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine." To quote

As for the second part, LiAlH4 is a strong reducing agent as it isn't in dry ether so the c=o becomes ch2. FYI RCO2H and LiAlH4 gives an alcohol but Carbonyl groups give the alkane right away with this.

Thanks

 

[studyingrobot457]

A condensation reaction occurs, so naturally if we have the same number of carbons as before, it's like a self condensing reaction so I suppose that's how the ring came to be.

"Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine." To quote

As for the second part, LiAlH4 is a strong reducing agent as it isn't in dry ether so the c=o becomes ch2. FYI RCO2H and LiAlH4 gives an alcohol but Carbonyl groups give the alkane right away with this.

Just a small correction, Alcohol is not produced because the amide group is being reduced, not a carobnyl

 

[mittul8yu98]

How was ur p4

 

[Kanekii]

Does any know the reason why incoming groups cannot be directed to 3 because of Methyl??

 

[A*****]

Does any know the reason why incoming groups cannot be directed to 3 because of Methyl??

I wrote that methyl is electron donating so other intermediates were more stable as the + charge was directly placed on the C atom next to methyl

 

[Mr."S"]

What were the values of delta H sol and the equilibrium 3