Here are some example combinations from my textbook:
In all of these combined products, the reactive centres seem to be the carbon at first position/ fourth position from the oxygen in priority. Is there something deep going on here?
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asked Aug 17, 2021 at 9:20
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3$\begingroup$ The 1' position on glucose is the anomeric carbon, which pretty much has to be involved because of the mechanism of the coupling reaction. The 4' position is common because it is opposite the 1' and leads to a linear polymer, but other positions (esp 6') are also very common. 6' is off the ring, so it allows for more flexibility in the linkage. 2' is quite sterically hindered. $\endgroup$– AndrewCommented Aug 17, 2021 at 12:35
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2$\begingroup$ Why is 2' hindered? @Andrew $\endgroup$– Cathartic EncephalopathyCommented Aug 18, 2021 at 4:18
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$\begingroup$ 2' is adjacent to 1', so if you have linkages at both positions (as in a polymer longer than two units), the two linked sugars interfere with each other. But it's not unknown. See Kojibiose and kojitriose, kojitetraose, etc. $\endgroup$– AndrewCommented Aug 18, 2021 at 13:43
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$\begingroup$ There is no need for a prime in these structures. Nucleotides have it to distinguish sugar atoms from nucleotide atoms. $\endgroup$– Karsten ♦Commented Aug 18, 2021 at 22:02
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