The above structure for questions 1-10 allows the President to have a full head
ID: 484671 • Letter: T
Question
The above structure for questions 1-10 allows the President to have a full head of hair. Unfortunately, it has side effects. The arrow is for question 9. How many carbons does the given compound have? 20 21 22 23 not a-d. How many methines does the given compound have? 8 7 6 5 not a-d. How many primary carbons does the given compound have? 5 6 7 8 not a.-d.. How many tertiary methines does the given compound have? 7 6 5 4 not a.-d. How many unsaturations are in the given compound? 3 4 5 5 not a.-d. How many unsaturated carbons are in the given compound? 6 5 4 3 not a.-d. Without considering resonance structures, the nitrogens in the given structure are? Linear octahedral tetrahedral trigonal not a.-d. What is the hybridization of the oxygens in the given structures? s sp sp^2 sp^2 not a.-d. What is the ideal absolute angle between two bonds to the atom indicate by the arrow in the given structure? 90degree 109.5degree 120degree 180degree nota.-d. How many hydrogens does the given compound have? 32 33 34 35 not a.-d.Explanation / Answer
1. d 23 carbons
2. Methine group is nothing but R1R2=C-H group This structure has only two such groups
so answer is d
3.
4. a tertiary methine is having C1C2C3= C-H group . There is no such group so ansewr d
5. one ring and one double bond is calculated as one unsaturation.
There are 4 rings and 3 double bond. so unsaturation is 4+3=7
6. 2 carbonyl and 2 in rings. so total 4 unsaturated carbons. so answer is c 4
7. In amide Amide nitrogen looks sp3 hybridized but it is sp2 because of conjugation
so if we ignore conjugation, it will sp3 and hence tetrahedral. so answer is c
8. carbonyl oxygen is generally sp2 hybridized so answer is c
9. b tetrahedral due to sp3 hybridization
10. there are 36 h so ansewr d
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