- Messages
- 112
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I think S in an amide then it is reduced to an amine.help ASAP
But the structure looks so funny.
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I think S in an amide then it is reduced to an amine.help ASAP
For biii) why do we divide the moles we got in bii) with 3?
the answer is so stupid
What are these stupid qns??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.Why doesn’t the second reaction occur ?
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 answerFor biii) why do we divide the moles we got in bii) with 3?
What is the ans?very very urgent help
Can you please send me the notes?all the topics in the syllabus bt that are sum notes to be learnt easily in pdf file
Why doesn’t the second reaction occur ?
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.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.
ThanksA 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.
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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