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Concept explainers
(a)
Interpretation:
The structure for each of the given name is to be drawn.
Concept introduction:
In case of rings containing seven or less than seven carbon atoms, the Z/E configuration is usually omitted for double bonds which are a part of the ring. The E configuration about the double bond is not stable due to ring strain. It is understood that the double bonded carbon atoms have a configuration such that they are cis to each other. In case of rings consisting of more than seven carbon atoms, the double bond, which is a part of the ring, can have E or Z configuration.
If the higher priority groups, attached to the double bonded carbon atoms, are on the same side of the double bond, the
(b)
Interpretation:
The structure for each of the given name is to be drawn.
Concept introduction:
In case of rings containing seven or less than seven carbon atoms, the Z/E configuration is usually omitted for the double bonds which are a part of the ring. The E configuration about the double bond is not stable due to ring strain. It is understood that the double bonded carbon atoms have a configuration such that they are cis to each other. In case of rings consisting of more than seven carbon atoms, the double bond, which is a part of the ring, can have E or Z configuration.
If the higher priority groups, attached to the double bonded carbon atoms, are on the same side of the double bond, the alkene is assigned Z configuration. If the higher priority groups, attached to the double bonded carbon atoms, are on the opposite side of the double bond, the alkene is assigned E configuration.
(c)
Interpretation:
The structure for each of the given name is to be drawn.
Concept introduction:
In case of rings containing seven or less than seven carbon atoms, the Z/E configuration is usually omitted for the double bonds which are a part of the ring. The E configuration about the double bond is not stable due to ring strain. It is understood that the double bonded carbon atoms have a configuration such that they are cis to each other. In case of rings consisting of more than seven carbon atoms, the double bond which is a part of the ring can have E or Z configuration.
If the higher priority groups, attached to the double bonded carbon atoms, are on the same side of the double bond, the alkene is assigned Z configuration. If the higher priority groups, attached to the double bonded carbon atoms, are on the opposite side of the double bond, the alkene is assigned E configuration.
(d)
Interpretation:
The structure for each of the given name is to be drawn.
Concept introduction:
In case of rings containing seven or less than seven carbon atoms, the Z/E configuration is usually omitted for the double bonds which are a part of the ring. The E configuration about the double bond is not stable due to ring strain. It is understood that the double bonded carbon atoms have a configuration such that they are cis to each other. In case of rings consisting of more than seven carbon atoms, the double bond which is a part of the ring can have E or Z configuration.
If the higher priority groups, attached to the double bonded carbon atoms, are on the same side of the double bond, the alkene is assigned Z configuration. If the higher priority groups, attached to the double bonded carbon atoms, are on the opposite side of the double bond, the alkene is assigned E configuration.
(e)
Interpretation:
The structure for the given name is to be drawn.
Concept introduction:
In case of rings containing seven or less than seven carbon atoms, the Z/E configuration is usually omitted for the double bonds which are a part of the ring. The E configuration about the double bond is not stable due to ring strain. It is understood that the double bonded carbon atoms have a configuration such that they are cis to each other. In case of rings consisting of more than seven carbon atoms, the double bond which is a part of the ring can have E or Z configuration.
If the higher priority groups, attached to the double bonded carbon atoms, are on the same side of the double bond, the alkene is assigned Z configuration. If the higher priority groups, attached to the double bonded carbon atoms, are on the opposite side of the double bond, the alkene is assigned E configuration.
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Chapter C Solutions
Organic Chemistry: Principles And Mechanisms
- Condensation polymers are produced when monomers containing two different functional groups link together with the loss of a small molecule such as H2O. The difunctional monomer H2N(CH2)6COOH forms a condensation polymer. Draw the carbon-skeleton structure of the dimer that forms from this monomer.arrow_forwardWhat is the structure of the monomer?arrow_forward→ BINDERIYA GANBO... BINDERIYA GANBO. AP Biology Notes Gamino acid chart - G... 36:22 司 10 ☐ Mark for Review Q 1 Hide 80 8 2 =HA O=A¯ = H₂O Acid HIO HBrO HCIO Question 10 of 35 ^ Σ DELL □ 3 % Λ & 6 7 * ∞ 8 do 5 $ 4 # m 3 ° ( 9 Highlights & Notes AXC Sign out Carrow_forward
- Which representation(s) show polymer structures that are likely to result in rigid, hard materials and those that are likely to result in flexible, stretchable, soft materials?arrow_forward3. Enter the molecular weight of the product obtained from the Williamson Ether Synthesis? OH OH & OH excess CH3l Ag₂Oarrow_forwardPlease answer 1, 2 and 3 on the endarrow_forward
- In the box below, specify which of the given compounds are very soluble in polar aprotic solvents. You may select more than one compound. Choose one or more: NaCl NH4Cl CH3CH2CH2CH2CH2CN CH3CH2OH hexan-2-one NaOH CH3SCH3arrow_forwardOn the following structure, select all of the atoms that could ACCEPT a hydrogen bond. Ignore possible complications of aromaticity. When selecting be sure to click on the center of the atom.arrow_forwardRank the compounds below from lowest to highest melting point.arrow_forward
- 18 Question (1 point) Draw the line structure form of the given partially condensed structure in the box provided. :ÖH HC HC H2 ΙΩ Н2 CH2 CH3 CH3 partially condensed formarrow_forwardsomeone else has already submitted the same question on here and it was the incorrect answer.arrow_forwardThe reaction: 2NO2(g) ⇌ N2O4(g) is an exothermic reaction, ΔH=-58.0 kJ/molrxn at 0°C the KP is 58.If the initial partial pressures of both NO2(g) and N2O4(g) are 2.00 atm:A) Is the reaction at equilibrium? If not, what is the value of Q? B) Which direction will the reaction go to reach equilibrium? C) Use an ICE table to find the equilibrium pressures.arrow_forward
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