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Chemistry: Post your doubts here!

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

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